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Find the curvature of $\mathbf{r}(t)=\left\langle t, t^{2}, t^{3}\right\rangle$ at the point

$(1,1,1) .$

$$

\kappa(1)=\frac{1}{7} \sqrt{\frac{19}{14}}

$$

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in this problem, gonna find the curvature of a vector are at a point. So here we're looking at the vector function composed of the three component functions T t squared and t cubed. We want to find the curvature at the 30.1 comma one, Apollo one. And if we look at the first coordinate in our vector and in the triple that were given, we see that we've got a pretty easy relationship between, uh, X, Y and Z and T, and we can say that t equals one corresponds to this point that we're interested in. So that's where we're gonna evaluate our curvature. I'm going to use their own one theorem 10 to find our curvature. So first thing we need to do is take the first derivative of our So our prime is going to be equal to one cheer t three t squared and our double prime our second derivatives. It's going to be equal to zero 26 t and then we want to take the crumbs product of those two, which we can write is the determinant of a three by three matrix. And if we have any rate that, um and I won't goes Well, yeah, If we evaluate that, we'll be looking at this two by two determinant times Vector I minus this two by two Determine at times vector j plus this two by two determinant time, Inspector Cate. And make sure that your fresh how to take the determinant of the three by three matrix if that's not familiar to you. So this is going to give us not gonna go ahead and write this in our angle bracket notation will end up with 12 t squared, minus 60 squared negative six T minus zero and tu minus zero as our three component functions, which is going to give us 60 squared negative 60 and two. Let's take the norm of that cross product to the norm of our prime cross with our double prime, which is going to give us a square root of each of our component functions squared. So 60 squared quantity squared is going to be 36 t to the fourth power plus negative 16 quantity of squared is going to be 36 t squared plus two squared, which is four. We can begin Ah, factor out a floor from that and then take the square root so we get up with it, too. Times the square root of nine T to the fourth power plus 90 squared plus one. So then we also going to find the norm of our prime so that what we haven't put in our denominator and okay is a square of each of these three component functions squared. So one squared or just one plus two t quantity squared or fourty squared plus nine t to the fourth power. And then, if you want Teoh here, then what we'll do is we'll evaluate, uh, or we'll come up with our function for the curvature using their intent. So and the numerator, we're gonna have to Times Square of 90 to the fourth power plus nine t squared plus one. And in our denominator, we're going to have the square root of one plus 40 squared, plus 90 to the fourth power, all raised to the third power and who want to evaluate that AT T equals one which we set up here. So what we're going to get is our numerator will have two times the square root of nine times one to the fourth power which is just one plus nine times T Square or nine times one squared, which is nine plus one minute. The denominator will have square root, have one plus four times T squared, just four times one squared or four plus nine times one to the fourth power, which is just nine. That's raised to the third power. This is gonna give us to Times Square to 19 all ever the square of 14 cubed. We know that the square root of 14 quantity cubed is the same thing as the square root of 14 Cube, which is I swear it in 2744. If, in my denominator I split that into the product of four and 686 I can factor the four or pull a four right out of the square root signs. We get your over square to 66 and when it's all said and done with, we can simplify this. So that way we get squared of 19 over seven Times Square to 14 and then at represents the curvature of the curve represented by our